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C.6.4 Integer programming

Let $A$ be an $ m \times n $ matrix with integral coefficients, $b\in Z\!\!\! Z^m$ and $c\in Z\!\!\! Z^n$. The problem

\begin{displaymath}\min\{c^T x \vert x\in Z\!\!\! Z^n, Ax=b, x\geq 0\hbox{ component-wise}\} \end{displaymath}

is called an instance of the integer programming problem or IP problem.

The IP problem is very hard; namely, it is NP-complete.

For the following discussion let $c\geq 0$ (component-wise). We consider $c$ as a weight vector; because of its nonnegativity, $c$ can be refined into a monomial ordering $>_c$. It turns out that we can solve such an IP instance with the help of toric ideals:

First we assume that an initial solution $v$ (i.e., $v\in Z\!\!\! Z^n, v\geq 0, Av=b$) is already known. We obtain the optimal solution $v_0$ (i.e., with $c^T v_0$ minimal) by the following procedure:

  • (1) Compute the toric ideal I(A) using one of the algorithms in the previous section.
  • (2) Compute the reduced Groebner basis G(c) of I(A) w.r.t. $>_c$.
  • (3) Reduce $x^v$ modulo G(c) using the Hironaka division algorithm. If the result of this reduction is $x^w$, then $w$ is an optimal solution of the given instance.

If no initial solution is known, we are nevertheless able to solve the problem with similar techniques. For this purpose we replace our instance by an extended instance with the matrix used in the Conti-Traverso algorithm. Indeed, the Conti-Traverso algorithm offers the possibility to verify solvability of a given instance and to find an initial solution in the case of existence (but none of the other algorithms does!). Details can be found in see [CoTr91] and see [The99].

An implementation of the above algorithm and some examples can be found in intprog_lib.

In general, classical methods for solving IP instances like Branch-and-Bound methods seem to be faster than the methods using toric ideals. But the latter have one great advantage: If one wants to solve various instances that differ only by the vector $b$, one has to perform steps (1) and (2) above only once. As the running time of step (3) is very short, solving all the instances is not much harder than solving one single instance.

For a detailed discussion see see [The99].